Problem: Chubby Yang is studying linear equations right now. He came up with a nice problem. In the problem you are given an n × n matrix W, consisting of integers, and you should find two n × n matrices A and B, all the following conditions must hold: 

  * Aij = Aji, for all i, j (1 ≤ i, j ≤ n); 
  * Bij = - Bji, for all i, j (1 ≤ i, j ≤ n); 
  * Wij = Aij + Bij, for all i, j (1 ≤ i, j ≤ n). 



Can you solve the problem?

Input

The first line contains an integer n (1 ≤ n ≤ 170). Each of the following n lines contains n integers. The j-th integer in the i-th line is Wij (0 ≤ |Wij| < 1717).

Output

The first n lines must contain matrix A. The next n lines must contain matrix B. Print the matrices in the format equal to format of matrix W in input. It is guaranteed that the answer exists. If there are multiple answers, you are allowed to print any of them.

The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 4.

Examples

Input

2
1 4
3 2


Output

1.00000000 3.50000000
3.50000000 2.00000000
0.00000000 0.50000000
-0.50000000 0.00000000


Input

3
1 2 3
4 5 6
7 8 9


Output

1.00000000 3.00000000 5.00000000
3.00000000 5.00000000 7.00000000
5.00000000 7.00000000 9.00000000
0.00000000 -1.00000000 -2.00000000
1.00000000 0.00000000 -1.00000000
2.00000000 1.00000000 0.00000000

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class CF{public static void main(String[] args){S s=new S();s.s();s.output();}}class S{
    void s() {
        int n = NI();
        double[][] W = new double[n][n];
        double[][] A = new double[n][n];
        double[][] B = new double[n][n];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                W[i][j] = NI();
        for (int i = 0; i < n; i++)
            for (int j = i; j < n; j++) {
                B[i][j] = (W[i][j] - W[j][i]) / 2;
                B[j][i] = B[i][j] == 0 ? 0 : -B[i][j];
                A[i][j] = A[j][i] = W[i][j] - B[i][j];
            }
        for (int i = 0; i < n; i++, sout())
            for (int j = 0; j < n; j++, out(' '))
                out(A[i][j]);
        for (int i = 0; i < n; i++, sout())
            for (int j = 0; j < n; j++, out(' '))
                out(B[i][j]);
    }


    BufferedReader stdin=new BufferedReader(new InputStreamReader(System.in));StringTokenizer st=new StringTokenizer("");final long LINF=Long.MAX_VALUE;
    StringBuilder output=new StringBuilder();final int INF=Integer.MAX_VALUE;void sout(Object x){output.append(x.toString()).append('\n');}
    void inLong(long[]a,int n){for(int i=0;i<n;i++)a[i]=NL();}int min(int i1,int i2){return i1<i2?i1:i2;}double ND(){return Double.parseDouble(NS());}
    long min(long i1,long i2){return i1<i2?i1:i2;}int max(int i1,int i2){return i1>i2?i1:i2;}long max(long i1,long i2){return i1>i2?i1:i2;}
    String NS(){while(!st.hasMoreTokens())try{st=new StringTokenizer(stdin.readLine());}catch(IOException ignored){}return st.nextToken();}
    int NI(){return Integer.parseInt(NS());}long NL(){return Long.parseLong(NS());}String NLn()throws IOException{return stdin.readLine();}
    int abs(int x){return x<0?-x:x;}long abs(long x){return x<0?-x:x;}void sout(){output.append('\n');}void out(Object x){output.append(x.toString());}
    int mod(int x,int mod){return(x+mod)%mod;}void output(){System.out.print(output);}void inInt(int[]a,int n){for(int i=0;i<n;i++)a[i]=NI();}
}